The Professor

[jayne]

my answer was that the professor's wife shook 4 peoples hands, is this what other people got too?

[AnswerGuy]

Geez Jayne, just blurt it out... but I agree, for what it's worth

[Cadmium]

Is this even possible?

[Pigboy]

What is wrong with having one person not shaking hands?

What I don't get is why the person who shakes hands with 4 people is fixed.



[This message has been edited by Pigboy (edited 05-15-2002 06:11 AM).]

[Cadmium]

I thought the quote I got from the riddle said that everbody shook hands. But if it is allowed for one person not to shake hands, it should be possible.

------------------
People who think they know everything really annoy those of us that do.

[Pigboy]

Sure, look at it in terms of pairs, The person who shakes 8 hands must be paired with the person who shakes with 0. Also with the 7 and 1 person. But why is it the Prof and wife who get 4 each and not some other arbitrary couple?

[Cadmium]

I was thinking pairs too, but now I read your post, I get it. If it was another couple, there would be two persons with the same amount of handshakes. The puzzle says that every answer was different.

------------------
People who think they know everything really annoy those of us that do.

[Pigboy]

I got ya, a bit sneaky, with emphasis on answers.

[Werebear]

...leave for eight hours, and two new puzzles come up, and get solved before I see them. *Grin*

Werebear

[Pigboy]

Yeah, think I've seen them before tho'.

[Keith]

Well its only 9 different answers cause the professor asked everyone ...he didnt ask himself....Right.

Soooo.....my answer is 4 !

[Icarus]

I'm going to be that annoying thorn in the side, and go back to Cadmium on this. This is impossible. The puzzle clearly states "each of the various guests and hosts greeted some of the others with handshakes".

Pigboy, that's what makes having one person not shake hands "wrong". The fact that someone could have refrained from shaking hands is eliminated by the use of the word "each". If 0 was allowed, it should have read "some".

[cha]

I agree with Icarus, but at the same time the most different people anyone could have shaken hands with is eight - so either:
1. The wording in the puzzle was in error,
2. Somebody lied,
3. Somebody brought a guest who left early, or
4. This all took place in a universe with more "flexible" mathematical laws.

[This message has been edited by cha (edited 05-15-2002 10:08 PM).]

[WildRain]

Well... Makes me think that one person could have shaken BOTH hands of another person. I can't offer an answer yet, but may be a direction... And what makes me think in that direction is a phraze: "how many DIFFERENT people's HANDS he/she had shaken"

[Pigboy]

Reckon the setter should just clarify if the question was mis-worded.

[fetus]

my question is why is it assumed that the person who shook 8 hands is paired with the person who shook 0 hands? lets say this happened (for hypothetical purposes):

i don't know anyone at this party. my wife/girlfriend/mistress/what-have-you know the prof and his wife. we shake only the hands of the people we don't know. thus i shake 8 hands and my wife/girlfriend/your-momma/what-have-you shakes 6. the same thing happens with all the other couples (one guy only knows 1 person, his wife/girlfriend/inflatable-doll/what-have-you knows 3, etc, etc) and it happens to work out the everyone knew just the right amount of other people so no one shakes the same # of hands (this, of course, doesn't solve the 9 answers/someone not shaking hands thing...). doing it this way takes out the ability to narrow down how many the prof's wife shook with, but you get the point. just seems like this is more faulty wording for the puzzle...

is there a step of logic i missed that implies my wife/girlfriend/dog/what-have-you and i shake 8 hands total, or are you all just assuming that because its the only way to make the puzzle work?

[fetus]

no wait!! i've got it!!!

(DISCLAIMER: this is called a JOKE. i don't need anyone yelling at me aboot ethics and other such nonsense. i haven't even been born yet... who am i to say women aren't people. talk aboot the pot calling the kettle gay...)

the puzzle is wrong because women aren't real people. thus, they cannot shake hands. their purpose is to prepare items to be placed ON the table, or items to be "serviced" UNDER the table. trick question. I'm on to you, grey labyrinth!

[edited disclamer]

[This message has been edited by fetus (edited 05-16-2002 11:27 AM).]

[mith]

Oops, bad wording on my part. It wasn't in the original. It has been fixed (I hope).

[Werebear]

Look at it this way. Everyone shook a different number of hands, not including the wife. SOMEONE shook 0, SOMEONE shook 1, 2, 3, 4, 5, 6, 7, 8. The wife has to repeat one of these numbers. OK?

Take the person with eight handshakes. they *HAD* to shake hands with everyone who shook hands... unless a person shakes hands with the person they came with, or shaking your own hand is applicable, in which case the puzzle is shot. *grin*
So one pairing HAS to exist - [8,0] - and the wife, by default, has shaken hands once.

Now take the person who shook 7 hands. They shook hands with everyone who shook hands except for their partner. If they were paired up with the 6, then you'd start repeating lower numbers... both would have had to shake the person who shook ONE hand. the only distribution that matches the criteria are: (and the * is the wife)
# - Shook hands with
8 - 7,6,5,4,3,2,1,*
7 - 8,6,5,4,3,2,*
6 - 8,7,5,4,3,*
5 - 8,7,6,4,*
4 - 8,7,6,5
3 - 8,7,6
2 - 8,7
1 - 8
0

If a person can't be paired with someone they shook hands with, it HAS to be [8,0] [7,1]... etc.

That make it any clearer?

Werebear

[fetus]

word

[Infinite|Insanity]

I don't understand. Why does the wife have to be the "*"? Why couldn't the wife be...say number "8"? Then she would have shaken 7 different people's hands. Have I completely missed something? Also, that would make the professor number "9" who shook hands with 8 other people.

[This message has been edited by Infinite|Insanity (edited 05-17-2002 05:01 PM).]

[Quailman]

The professor got 8 different responses.

The most anyone could have shaken is 8 and the least is 0, but only if that person is the spouse of Mr. or Mrs. 8 (everyone else had at least one, because Mr. 8 shook them all.

The next most is 7, skipping one person (Mrs. 8). Mr. 7 is married to Mrs. 1 because now everyone else has been greeted by 7 & 8, for a minimum of 2. Six is married to 2 and 5 married to three. So you see, the total of each couple must be 8, and there are two who shook 4 hands each. If they were anyone but The Professor and his wife, there wouldn't be eight different responses to the question.

[cha]

but there were 9 different responses... Except fot that, good explanation.

[Quailman]

Right. Typo on my part. 9 different responses - Zero through Eight.

[Carlos Beek]

the wife, the profesor and all the people can shake 8 hands, to recive 9 diferent answers to the question of the profesor, the answer is:
guest1 = 8 hands
guest2 = 7 hands
guest3 = 6 hands
guest4 = 5 hands
wife & profesor = 4 hands each
guest5 = 3 hands
guest6 = 2 hands
guest7 = 1 hands
if you made the grafic whith these numbers you can verify the numbers!!
guest8 = 0 hand.

[Rmselby7]

I'm confused because it says that he asked each OTHER person. There are 9 people so there are 4 or 5 OTHER people. How can you get 9 different answers out of only 4 or 5 people? Please explain.

My answer, by the way everyone is talking, would have to be 0-8 because it doesn't clarify which person the wife is. She could be the first who answered 8 or the last who answered 0.

[The Seventh]

But this doesn't make sense as to why they would get 9 different responses. Ok, there are ten people. Disregarding the professor, since he's asking the question, there are 9. Now, the male person in each couple could not shake hands with the female person in each couple, and vice versa. That in mind, wouldn't each member of the couple get the same results?

For example, the husband of Couple A would have shaken hands with the Professor, his wife, both members of couple B, both members of couple C, and both members of couple D. The wife of couple A could have done the exact same thing. So does this mean that if the husband of couple A shook hands with someone, the wife of couple B did not?

[jjeffrey]

The solution of 4 is obvious when you reduce the problem.
Claim 1: Ignoring the Prof each person shook between 0 and 8
hands, and nobody shook the same amount (all 9 numbers are represented).
Claim 2: The person who shook 8 hands is married to the person who
shook 0 hands. This cannot be the Prof & wife, as this would
violate claim 1.
Claim 3: Remove the 8-0 couple from the problem, and you get an equivalent
reduced problem. This new problem involves 4 couples, and ignoring the prof,
everybody shook between 0 and 6 hands (all 7 numbers represented). Solve this new problem, and add 1 shake to each persons total to solve the original problem.
Claim 4: To solve this new problem, repeat Claims 2 & 3 to reduce the problem further. Remove the 6-0 couple, and now solve this problem. Add 2 handshakes for final answer.
Claim 5: Reduce again. Remove the 4-0 couple. Now the problem is getting
simple. We are now left with two couples, 4 people. Ignoring the prof, somebody shook 0 hands, somebody 1, and somebody 2 hands. This is easy to solve, but why solve it when you can reduce again?
Claim 6: Reduce again. Get rid of the 2-0 couple, and now you are left
with a trivial problem. There is one couple, and ignoring the prof, somebody
shook 0 hands (I'll bet it was his wife!).
Claim 7: We reduced the problem 4 times, so to get back to the original problem, just recurse back 4 times, adding one handshake to each person
at each level. This gives 4 handshakes for the wife.

[The Seventh]

But what keeps the husband of a couple from shaking hands with the same person as the wife of a couple? In reality, by matching up pairs with a corresponding number (8-0, 7-1, 6-2, etc.) we're assuming that no member of a couple shook hands with the same person as the other. That's not stated within the problem. In reality, there's nothing stopping 8 from being paired with 2 or 7 or anything other than 0.

[CrystyB]

Seventh, the wife could know somebody/some couple that the husband did not. Here's a possible scenario, using the reduced solution that i just read:

code:

---

   Ah Aw Bh Bw P  PW


Ah .. .. ** ** ** **


Aw .. .. -- -- -- --


Bh ** -- .. .. ** **


Bw ** -- .. .. -- --


P  ** -- ** -- .. ..


Pw ** -- ** -- .. ..

---

Notice the B couple (1-3). The numbers add up to 4, but that does not mean the sets of handshake partners are complementary to one another...

[The Seventh]

Granted, but what keeps the wife from shaking more than one hand?

[CrystyB]

Let's start all over again. Suppose wlog each husband shakes more hands than his wife. Name the 4 one "A". Now his wife is the only remaining choice for the 0. Anyone else had at least one handshake (with Ah). So 4-0. Name the 3 one "B". Now B's wife is the only choice for the 1. That leaves the Professor and his wife with 2 handshakes each.

edit: had the professor been an already labeled husband, this last couple would have reported 2 each, thus the professor had no 5 different answers.

[This message has been edited by CrystyB (edited 05-19-2002 09:06 PM).]

[The Seventh]

Oh, now I get what you're saying. Ok. Thanks CrystyB. Yeah, the answer would then have to be 4.

[Werebear]

Inifinite Insanity - the * merely was a place marker for handshakes unaccounted for. In other words, for a person to shake hands with 8 people, they'd have to shake hands with the person who shook 1,2,3,4,5,6,7... and SOMEONE. Of course, there's only one person unaccounted for, the SOMEONE has to be the wife.

I hope this makes things clearer.

Werebear

[proximity]

suppose that the professor himself didn't shake any hands.
then the most hands that any one person cold have shook would be 7
(10 people minus themself, minus their SO, minus the professor)

EXCEPT for the professor's wife. The most hands that she could have shook would be 8
(10 people minus the professor/her SO)

so to get 9 different answers,

wife shakes 8
someone shakes 7
someone shakes 6
someone shakes 5
someone shakes 4
someone shakes 3
someone shakes 2
someone shakes 1

but that doesn't work because that isn't 9 people ... so you would need a 9th person. someone who
a.) shook the same number of hands as someone else
b.) shook 9 hands
c.) shook 0 hands

a.) doesn't work because that's the question ...
b.)doesn't work because they can't shake their own hand or that of their SO
c.) so then ...

professor shakes 0
wife shakes 7
someone shakes 6
someone shakes 5
someone shakes 4
someone shakes 3
someone shakes 2
someone shakes 1
someone shakes 0

and we still have the same original problem ... so that doens't help us to have a person who shakes no hands.


so, now assume that the professor is shakes 8 hands....
then his wife can shake any number of hands, from 0 to 8 ...
wife shakes 8
someone shakes 7
someone shakes 6
someone shakes 5
someone shakes 4
someone shakes 3
someone shakes 2
someone shakes 1
someone shakes 0

so that works, but the wife could have been any of those people, AND you have to use the 0 thing.

note: it is not possible for both the prof and his wife to shake no hands because it says that various guests and HOSTS greeted some of the others with handshakes.

the only thing that can distinguish the professor's wife from the other guests (and make it so that you can tell how many hands SHE shook) is the professor himself.
because she can't shake hands with him.... so if he shook hands with no one ... then she would be the _only_ person who could have shaken hands with 8 people...
so the only way that I can see this working is
professor = 0
wife = 8
someone = 7
someone = 6
someone = 5
someone = 4
someone = 3
someone = 2
someone = 1
someone = 0

two people have shaken no hands ... the prof and one of the guests, but he can still get 9 different answers to his question because he can't ask himself. ^_^
that's what I think the solution is.

[proximity]

whoops.
but the wife can't shake 8 hands if one of those losers out there didn't shake hands with anyone ...
damn
nevermind I guess.
erk.

[Werebear]

...the professor interviewed everyone else but himself and his wife. He knows how many handshakes he gave.

He got nine different answers. They have to be - 0,1,2,3,4,5,6,7,8. Yes?

So someone shook eight hands. It could have been the professor for all we know at the moment... but let's see.
SOMEONE shook 8 hands... the hands of 1,2,3,4,5,6,7... AND someone not interviewed. Who was not interviewed? The professor's wife. so SOMEONE shook the hands of 1-7, and the wife. it could NOT have been the professor, he wouldn't have shaken his wife's hands. Yes? So the professor was NOT the person who shook 8 hands, or their partner.

ALSO, whoever shook 8 hands shook everyone's hand but the person with ZERO handshakes, so the person with zero handshakes HAS to be their partner. Yes?

So we have a set. [8 handshakes and 0 handshakes]... and they're not the professor or his wife. We do know that the wife HAS recieved one handshake so far... the person with eight handshakes shook her hand. ALSO, the person with one handshake is done... no more handshakes for them. The only person they shook hands with is the person with 8 handshakes. Yes?

SO NOW. We have people who shook 1,2,3,4,5,6,7 (all who have shaken hands once, with 8) Gee.. if you subtract the one handshakes from each person, their numbers of handshakes left is 0,1,2,3,4,5,6. THIS looks familiar. anyways, the person with 7 handshakes shook hands with 8 (of course) They didn't shake hands with 0 or 1 (of course - they're done.) That leaves the list of people whose hands they could shake at 2,3,4,5,6,wife. IN ORDER TO HAVE 7 HANDSHAKES, 7 MUST HAVE SHAKEN EVERYONE ON THIS LIST'S HAND... so their partner HAS to be 1, the only person left unpaired whose hand they CAN'T shake. So 7 shakes the hands of 2,3,4,5,6,wife,8. Their partner is 1. They can't be the professor or the wife. Yes?

SO... now we have [8,0] and [7,1]. The wife has received two handshakes. The list of people left is 2,3,4,5,6. Each have received two handshakes. (one from 7, one from 8) Gee.. if you subtract two handshakes from each person, their numbers of handshakes left is 0,1,2,3,4. THIS looks familiar. Anyways, the person with 6 handshakes shook hands with 7&8 (of course). They didn't shake hands with 0,1, or 2 (of course - they're done) This leaves the list of people whose hands they could shake at 3,4,5,wife. IN ORDER TO HAVE 6 HANDSHAKES, 6 MUST HAVE SHAKEN EVERYONE ON THIS LIST'S HAND... so their partner has to be 2, the only person left unpaired whose hand they CAN'T shake. So 6 shakes the hand of 3,4,5,7,8,wife. Their partner is 2. they can't be the professor or his wife. Yes?

So... now we have [8,0], [7,1] and [6,2]. The wife has received three handshakes. The list remaining is 3,4,5. Let's just assume we can do the same thing for [5,3], as it's the same scenario. (subtracting three handshakes gives 0,1,2 - 5 shakes the hands of 4,6,7,8,wife, 3 has to be the partner) gaining the wife another handshake, exactly the same as the previous two examples.

SO. We have these sets. [8,0] [7,1] [6,2] [5,3]. That leaves 4 unmatched... the professor. In the process, the wife has also gathered four handshakes.

Chart (again)
# - person number
P - partner
W - wife

# - P - Shook Hands With
8 - 0 - 1,2,3,4,5,6,7,w
7 - 1 - 2,3,4,5,6,8,w
6 - 2 - 3,4,5,7,8,w
5 - 3 - 4,6,7,8,w
4 - w - 5,6,7,8 - - The Professor!
w - 4 - 5,6,7,8 - - The Wife!
3 - 5 - 6,7,8
2 - 6 - 7,8
1 - 7 - 8
0 - 8

[Borodog]

I still say it was Gilligan.


------------------
Insert humorous sig here.

[The Professor]

GILLIGAN! DROP THOSE COCONUTS! #BONK# Owwwwww!

[BigBunch]

Something no one has mentioned: the professor and his wife not only shake the same number of hands, but they shake the same persons' hands. not only that, but also, every person the prof and wife shake hands with id of the same sex: either all 4 men or all 4 women. I think we need to examine the sex life of the professor and his wife.

Is the prfessor gay and trying to pick up the men? If that was the case, I am sure the wife would be very sexually frustrated and also be trying to pick up the men.

Maybe the wife is gay and trying to pick up the women. In this case the professor himself is the frustrated one.

Or is could be that either the professor or his wife is bisexual, and they want a threesome. Since this is a dream world, I will assume the wife is the bisexual one. I do have to admit though that I have watched enough Gilligan's Island to have my doubts about the professors sexual orientation.